Question

A projectile is thrown with speed $u$ making angle $\theta$ with horizontal at $t=0.$ It just crosses two points of equal height, at time $t=1\text{s}$ and $t=3\text{s}$ respectively. Calculate the maximum height attained by it?
$(g=10{\text{m/s}}^2)$

• A. $5\text{m}$
• B. $10\text{m}$
• C. $15\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is D $20\text{m}$


Given:
$t_P=1\text{s};t_Q=3\text{s}$

For point $\text{P or Q}:$

Using 2nd equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 2h=2u_y.\left(t\right)-g{t}^2$

$\Rightarrow 5t^2-u_y.\left(t\right)+h=0$

Let roots be $t_1=1$ and $t_1=3$, then

$t_1+t_2=\frac{-(-u_y)}{5}$

$1+3=\frac{u_y}{5}$

$\Rightarrow u_y=20\text{m/s}$

Maximum height, $h_{max}=\frac{u_y^2}{2g}=\frac{{20}^2}{20}=20\text{m}$

Hence, option $\left(d\right)$ is correct.